GCSE Maths Coursework Growing Shapes

My aim is to investigate and explain the growth of congruent shapes by using different methods. I am going to try and find out the formulae for the following in this order:

Length

Perimeter

Area

Number of shapes

Number of Lines (inner, outer)

Number of shapes added

Number of vertices

At the end of my investigation I am going to look at my results for both squares and triangles and compare them to see if they have anything in common or are similar in any way.

I also hope to extend my coursework and look at 3D shapes or other shapes such as pentagons, hexagons etc…

I am going to work systematically and look to spot any patterns, and in the end, establish an algebraic formula for all things listed above.

The patterns for squares are as follows

1234

The patterns for triangles are as follows

1234

5

The pattern for squares grows by adding another square to each face every time:

The pattern for triangles grows by adding another triangle to each face every time:

Growing Shapes: Squares

Perimeter

Pattern no. (n)

Perimeter

1

4

2

12

3

20

4

28

5

36

D1

As there are all 8’s in the D1 column, the formula contains 8n

Pattern no. (n)

Perimeter

Perimeter – 8n

1

4

-4

2

12

-4

3

20

-4

4

28

-4

5

36

-4

Formula for perimeter of squares – Formula = 8n-4

Check

When n = 5

Perimeter = 8n - 4

= 8 × 5 – 4

= 36

Length and Width

Pattern no. (n)

Length

1

1

2

3

3

5

4

7

5

9

D1

As there are all 2’s in the D1 column, the formula contains 2n

Pattern no. (n)

Length

Length – 2n

1

1

-1

2

3

-1

3

5

-1

4

7

-1

5

9

-1

Formula for length of squares – formula = 2n – 1

Check

When n = 3

Length = 2n – 1

= 2 × 3 – 1

= 5

Number of Squares/Area

Pattern no. (n)

No. of Squares

1

1

2

5

3

13

4

25

5

41

D1 D2

As there are all 4’s in the D2 column,the formula contains 2n2.

Pattern no. (n)

No. of Squares

No. of Squares – 2n2

1

1

-1

2

5

-3

3

13

-5

4

25

-7

5

41

-9

D1

As there are all -2’s in the D1 column, the formula contains 2n2 – 2n.

I need to substitute in the pattern number to find out what to do next.

I am going to use 3 to start with.

(2×32) – (2×3) = 12

To get from 12 to 13, I need to add 1.

I am now going to substitute in 4 to see if it works as well.

(2×42) – (2× 4) = 24

To get from 24 to 25, I need to add 1 so the formula = 2n2 – 2n + 1 and works for everything.

Check

When n = 2

Area = 2n2 -2n + 1

= 2×22 – 4 +1

= 5

I have to use another method to find the formula so I am going to use the odd numbers method.

I notice that with pattern 4 in columns when n = 4:

2 lots of (1 + 3 + 5) = 2 x 9

1 lot of 7 = 1 x 7

2 x 9 = 2(n-1)2

7 = 2n – 1

Ts = 2(n-1)2 + 2n – 1

Number of Lines

Pattern no. (n)

No. of lines

1

4

2

16

3

36

4

64

5

100

D1 D2As there are all 8’s in the D2 column, the formula contains 4n2.

Pattern no. (n)

No. of lines

No. of lines - 4n2

1

4

0

2

16

0

3

36

0

4

64

0

5

100

0

The formula for number of lines = 4n2

Check

Number of lines = 4n2

= 4 × 32

= 36

Number of Shapes Added

I started with pattern number 2 because there is no previous shape to add the squares onto

Pattern no. (n)

No. shapes added

2

4

3

8

4

12

5

16

6

20

D1

As there are all 4’s in the D1 column, the formula contains 4n.

Pattern no. (n)

No. shapes added

No. shapes added - 4n

2

4

-4

3

16

-4

4

36

-4

5

64

-4

6

100

-4

The formula for number of shapes added = 4n – 4

Check

Number of shapes added = 4n – 4

= 4 × 3 – 4

= 8

Number of outer vertices

Pattern no. (n)

No. of outer vertices

1

4

2

8

3

12

4

16

5

20

D1

As there are all 4’s in the D1 column, the formula contains 4n.

Pattern no. (n)

No. of outer vertices

No. of outer vertices - 4n

1

4

0

2

8

0

3

12

0

4

16

0

5

20

0

Formula for number of outer vertices = 4n

Check

Number of outer vertices = 4n

= 4 × 3

= 12

Growing Shapes: Triangles

Perimeter

Pattern no. (n)

Perimeter

1

3

2

6

3

12

4

15

5

21

6

24

7

30

D1 D2 D3

I can tell that this method for working out the perimeter does not work because the numbers in the difference columns alternate between positive and negative.

Instead of trying to find a formula for the all of the triangular patterns, I can try and find 2 separate formulas for the patterns where the added triangles are pointing up and the patterns where the triangles are pointing down.

I have targeted the streets that have in my opinion serious issues and in some cases, like that of Bounty Road, I have pin pointed a good value to the area to explain why I have found results like I have.

of one half, I will need to do: Base of one triangle = 400 / 2 = 200m Now, I will use the Pythagoras Theorem to calculate length A (the height of the triangle): Length A = c2 - b2 = a2 = 3002 - 2002 = 500002 = V50000 = 223.60m Therefore, the height of the triangle is 223.607m.

By using 21cm as the length, the maximum area of the trapezium is greater than the triangular and rectangular cross sectional areas but is still less than the semicircle therefore my hypothesis is still true. 7cm 7cm (L/3) (L/3) 7cm (L/3)

maximum area of the semi-circle is 143.2cm� Half-octagon I will be using half of a regular octagon. There is nothing to vary in a regular octagon as all the sides and angles are equal, so I will only be getting one value for the half-octagon shaped guttering.

I can now work out its area: Area = 1/2 base x height = (1/2 x 350) x 273.9 = 175 x 273.9 = 47925.7m� Spreadsheet: I am now going to obtain more results using a spreadsheet-software and draw a graph out of it.